For what values of x: left[begin{array}{lll}1 | vedclass

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(A) We have $\left[\begin{array}{lll}1 & 2 & 1\end{array}ight]\left[\begin{array}{lll}1 & 2 & 0 \ 2 & 0 & 1 \ 1 & 0 & 2\end{array}ight]\left[\be

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(A) We have $\left[\begin{array}{lll}1 & 2 & 1\end{array}ight]\left[\begin{array}{lll}1 & 2 & 0 \ 2 & 0 & 1 \ 1 & 0 & 2\end{array}ight]\left[\begin{array}{l}0 \ 2 \ x\end{array}ight]=0$. First,multiply the first two matrices: $\left[\begin{array}{lll}1 & 2 & 1\end{array}ight]\left[\begin{array}{lll}1 & 2 & 0 \ 2 & 0 & 1 \ 1 & 0 & 2\end{array}ight] = \left[\begin{array}{lll}1(1)+2(2)+1(1) & 1(2)+2(0)+1(0) & 1(0)+2(1)+1(2)\end{array}ight] = \left[\begin{array}{lll}6 & 2 & 4\end{array}ight]$. Now,multiply the result by the third matrix: $\left[\begin{array}{lll}6 & 2 & 4\end{array}ight]\left[\begin{array}{l}0 \ 2 \ x\end{array}ight] = 6(0) + 2(2) + 4(x) = 0 + 4 + 4x = 4 + 4x$. Setting the result to $0$: $4 + 4x = 0 \Rightarrow 4x = -4 \Rightarrow x = -1$. Thus,the required value of $x$ is $-1$.

For what values of $x$: $\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=O$?

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